Pre-Calculus Homework Solutions 217

# Pre-Calculus Homework Solutions 217 - sin Q q sin 110.7...

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13. } sin q Q } 5 ¬ } sin r R } } sin 1 1 7 1 . 0 2 .7° } 5 ¬ } s 9 in .8 R } 9.8sin110.7° 5 ¬ 17.2sin R } 9.8si 1 n 7 1 .2 10.7° } 5 ¬ sin R sin 2 1 1 } 9.8si 1 n 7 1 .2 10.7° } 2 5 ¬ R 32° < ¬ R m / P 1 m / Q 1 m / R 5 ¬ 180 m / P 1 110.7 1 32 < ¬ 180 m / P 1 142.7 < ¬ 180 m / P < ¬ 37 } sin p P } 5 ¬ } sin q Q } } sin p 37° } 5 ¬ } sin 1 1 7 1 . 0 2 .7° } 17.2sin37° 5 ¬ p sin110.7° } 1 s 7 in .2 1 s 1 in 0. 3 7 7 ° ° } 5 ¬ p 11.1 < ¬ p 14. A w D w i B w C w , so / DAC > / BCA because they are alternate interior angles. Thus, m / DAC 5 88. } sin 6 32° } 5 ¬ } si D n8 C } DC sin32° 5 ¬ 6sin88° DC 5 ¬ } 6 s s in in 3 8 2 8 ° ° } DC < ¬ 11.3 ABCD is a parallelogram, so A w D w > B w C w and D w C w > A w B w . The perimeter of ABCD is 2(6) 1 2(11.3), or 34.6 units. 15. m / A 1 m / B 1 m / C 5 ¬ 180 55 1 m / B 1 62 5 ¬ 180 m / B 1 117 5 ¬ 180 m / B 5 ¬ 63 } si 2 n 4 6 0 } 5 ¬ } sin A 6 B } AB sin63° 5 ¬ 240sin62° AB 5 ¬ } 24 s 0 in si 6 n 3
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